∀: ((∀ ∶ ( = ) ∨ ( ⊈ )) → ( = ∅)) where A and B are sets.
The main problem was to either prove or disprove the statement.
I do not know how to approach the proof.
If I fix A, then if ∀ B ((B=A) ∨ (B ⊈ A)) is false, any implication from it always true so we can ignore the cases when it is false. Because P→Q is always true when P is false.
Now, about the case when for a fix A , ∀ B ((B=A) ∨ (B ⊈ A)) is true. What should it imply? Because when P is true and Q is false P→Q is false.
I am stuck here and I don't even know if this is the way to solve. Please show a proper approach.
Thank you.
There are a few points to make here.
First, if you're trying to prove an implication $P \to Q$, there's no need to do case analysis on whether $P$ is true or false. This just unnecessarily lengthens the proof. Instead, simply assume $P$ and prove $Q$.
So in this case, we assume that $\forall B . (B = A \lor B \nsubseteq A)$. We seek to prove $A = \emptyset$.
As suggested in the comments, the correct thing to do here is to set $B = \emptyset$ (or, more formally, apply the universal elimination rule on $B = \emptyset$). Then we see that either $\emptyset = A$ or $\emptyset \nsubseteq A$. I'll let you finish the proof from here.
If we had significant difficulty proving $A = \emptyset$ from the hypothesis $\forall B . (B = a \lor B \nsubseteq A)$, we would have tried to come up with a counterexample.